A visual introduction to machine learning by Yee and Chu.
Yee and Chu created a step-by-step build of a recursive binary tree to model the differences between homes in SF and homes in NYC. http://www.r2d3.us/visual-intro-to-machine-learning-part-1/
Classification and Regression Trees (CART)
Basic Classification and Regression Trees (CART) Algorithm:
Start with all observations in one group.
Find the variable/split that best separates the response variable (successive binary partitions based on the different predictors / explanatory variables).
Evaluation “homogeneity” within each group
Divide the data into two groups (“leaves”) on that split (“node”).
Within each split, find the best variable/split that separates the outcomes.
Continue until the groups are too small or sufficiently “pure”.
Prune tree.
Minimize heterogeneity
For every observation that falls into the region \(R_m\),
prediction = the mean of the response values for observations in \(R_m\).
\(\Rightarrow\) Minimize Residual Sum of Squares (RSS): \[RSS = \sum_{m=1}^{|T|} \sum_{i \in R_m} (y_i - \overline{y}_{R_m})^2\] where \(\overline{y}_{R_m}\) is the mean response for observations within the \(m\)th region.
Recursive binary splitting
Select the predictor \(X_j\) and the cutpoint \(s\) such that splitting the predictor space into the regions \(\{X | X_j< s\}\) and \(\{X | X_j \geq s\}\) lead to the greatest reduction in RSS.
For any \(j\) and \(s\), define the pair of half-planes to be \[R_1(j,s) = \{X | X_j < s\} \mbox{ and } R_2(j,s) = \{X | X_j \geq s\}\] Find the value of \(j\) and \(s\) that minimize the equation: \[\sum_{i:x_i \in R_1(j,s)} (y_i - \overline{y}_{R_1})^2 + \sum_{i:x_i \in R_2(j,s)} (y_i - \overline{y}_{R_2})^2\]
where \(\overline{y}_{R_1}\) is the mean response for observations in \(R_1(j,s)\) and \(\overline{y}_{R_2}\) is the mean response observations in \(R_2(j,s)\).
Trees in action
Measures of impurity
\(\hat{p}_{mk}\) = proportion of observations in the \(m\)th region from the \(k\)th class.
classification error rate = fraction of observations in the node & not in the most common class:
cross-entropy\[D_m = - \sum_{k=1}^K \hat{p}_{mk} \log \hat{p}_{mk}\] (Gini index & cross-entropy will both take on a value near zero if the \(\hat{p}_{mk}\) values are all near zero or all near one.)
Recursive binary splitting
For any \(j\) and \(s\), define the pair of half-planes to be \[R_1(j,s) = \{X | X_j < s\} \mbox{ and } R_2(j,s) = \{X | X_j \geq s\}\]
Seek the value of \(j\) and \(s\) that minimize the equation: \[\begin{align}
& \sum_{i:x_i \in R_1(j,s)} \sum_{k=1}^K \hat{p}_{{R_1}k}(1-\hat{p}_{{R_1}k}) + \sum_{i:x_i \in R_2(j,s)} \sum_{k=1}^K \hat{p}_{{R_2}k}(1-\hat{p}_{{R_2}k})\\
\\
\mbox{equivalently: } & \\
& n_{R_1} \sum_{k=1}^K \hat{p}_{{R_1}k}(1-\hat{p}_{{R_1}k}) + n_{R_2} \sum_{k=1}^K \hat{p}_{{R_2}k}(1-\hat{p}_{{R_2}k})\\
\end{align}\]
Stopping
We can always make the tree more “pure” by continuing the split.
Too many splits will overfit the model to the training data!
There is a cost to having a larger (more complex!) tree.
Define the cost complexity criterion, \(\alpha > 0:\)\[\begin{align}
\mbox{numerical: } C_\alpha(T) &= \sum_{m=1}^{|T|} \sum_{i \in R_m} (y_i - \overline{y}_{R_m})^2 + \alpha \cdot |T|\\
\mbox{categorical: } C_\alpha(T) &= \sum_{m=1}^{|T|} \sum_{i \in R_m} I(y_i \ne k(m)) + \alpha \cdot |T|
\end{align}\] where \(k(m)\) is the class with the majority of observations in node \(m\) and \(|T|\) is the number of terminal nodes in the tree.
\(\alpha\) small: If \(\alpha\) is set to be small, we are saying that the risk is more worrisome than the complexity and larger trees are favored because they reduce the risk.
\(\alpha\) large: If \(\alpha\) is set to be large, then the complexity of the tree is more worrisome and smaller trees are favored.
In practice
Consider \(\alpha\) increasing. As \(\alpha\) gets bigger, the “best” tree will be smaller.
The test error will not be monotonically related to the size of the training tree.
A note on \(\alpha\)
In the text (Introduction to Statistical Learning) and almost everywhere else you might look, the cost complexity is defined as in previous slides.
However, you might notice that in R the cost_complexity value is typically less than 1. From what I can tell, the value of the function that is being minimized in R is the average of the squared errors and the missclassification rate.
Use recursive binary splitting to grow a large tree on the training data, stopping only when each terminal node has fewer than some minimum number of observations.
Apply cost complexity pruning to the large tree in order to obtain a sequence of best subtrees, as a function of \(\alpha\).
Use \(V\)-fold cross-validation to choose \(\alpha\). Divide the training observations into \(V\) folds, then for each \(v=1, 2, \ldots, V\):
Repeat Steps 1 and 2 on all but the \(v\)th fold of the training data.
Evaluate the mean squared prediction error on the data in the left-out \(k\)th fold, as a function of \(\alpha\). For each value of \(\alpha\), average the prediction error (either misclassification or RSS), and pick \(\alpha\) to minimize the average error.
Return the subtree from Step 2 that corresponds to the chosen value of \(\alpha\).
# A tibble: 8 × 4
variable type role source
<chr> <list> <chr> <chr>
1 island <chr [3]> predictor original
2 bill_length_mm <chr [2]> predictor original
3 bill_depth_mm <chr [2]> predictor original
4 flipper_length_mm <chr [2]> predictor original
5 body_mass_g <chr [2]> predictor original
6 sex <chr [3]> predictor original
7 year <chr [2]> predictor original
8 species <chr [3]> outcome original
Test and training error as a function of model complexity. Note that the error goes down monotonically only for the training data. Be careful not to overfit!! image credit: ISLR
